Asked by Anonymous
Find how many integers between 200 and 500 are divisible by 8
Answers
Answered by
Arora
You can solve this problem as an Arithmetic Progression.
The series: 208, 216, 224,...., 488, 496
a(n) = 496
a(1) = 208
d = 8
For an A.P.,
a(n) = a(1) + (n-1)d
=> 496 = 208 + (n-1)8
=> 288/8 = n - 1
=> n = 36 + 1 = 37
Hence, there are 37 such integers.
The series: 208, 216, 224,...., 488, 496
a(n) = 496
a(1) = 208
d = 8
For an A.P.,
a(n) = a(1) + (n-1)d
=> 496 = 208 + (n-1)8
=> 288/8 = n - 1
=> n = 36 + 1 = 37
Hence, there are 37 such integers.
Answered by
Bosnian
Find the first number in thist interval that is divisible by 8
In this case:
200 / 8 = 25
Find the last number in thist interval that is divisible by 8
In this case:
496 / 8 = 62
Numbers 200 and 496 assume arithmetic progression, where first member a1 = 200 and last member a62 = 496
Common difference is d = 8
In arithmetic progression:
an = a1 + ( n - 1 ) d
In this case:
a62 = a1 + ( n - 1 ) d
496 = 200 + ( n - 1 ) ∙ 8
496 = 200 + 8 n - 8
496 = 192 + 8 n
Subtract 192 to both sides
496 - 192 = 192 + 8 n - 192
304 = 8 n
Divide both sides by 8
38 = n
n = 38
In this case:
200 / 8 = 25
Find the last number in thist interval that is divisible by 8
In this case:
496 / 8 = 62
Numbers 200 and 496 assume arithmetic progression, where first member a1 = 200 and last member a62 = 496
Common difference is d = 8
In arithmetic progression:
an = a1 + ( n - 1 ) d
In this case:
a62 = a1 + ( n - 1 ) d
496 = 200 + ( n - 1 ) ∙ 8
496 = 200 + 8 n - 8
496 = 192 + 8 n
Subtract 192 to both sides
496 - 192 = 192 + 8 n - 192
304 = 8 n
Divide both sides by 8
38 = n
n = 38
Answered by
Arora
200 shouldn't be included, should it?
The question says 'between 200 and 500', so.
The question says 'between 200 and 500', so.
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